Problem: What is the value of the following logarithm? $\log_{2} \left(\dfrac{1}{8}\right)$
Explanation: If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $2^{y} = \dfrac{1}{8}$ In this case, $2^{-3} = \dfrac{1}{8}$, so $\log_{2} \left(\dfrac{1}{8}\right) = -3$.